Let $\mathbb{R}^{2}$ be a normed space with $\|(x, y)\|_{1}=|x|+|y|$ and $f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ defined by : $$ f(x, y)=\left(\frac{1}{4} \sin (x+y), 1+\frac{2}{3} \arctan (x-y)\right) $$
- Show that there exists $k \in] 0,1\left[\right.$ such that, for all $(x, y),\left(x^{\prime}, y^{\prime}\right) \in \mathbb{R}^{2},$ we have $$ \left\|f(x, y)-f\left(x^{\prime}, y^{\prime}\right)\right\|_{1} \leq k\left\|(x, y)-\left(x^{\prime}, y^{\prime}\right)\right\|_{1} $$
- Deduce that $$ \left\{\begin{aligned} \frac{1}{4} \sin (x+y) &=x \\ 1+\frac{2}{3} \arctan (x-y) &=y \end{aligned}\right. $$ admits a unique solution in $\mathbb{R}^{2}$.
- can we use the same method using the norm $\|.\|_{\infty}$ instead of $\|\cdot\|_{1}$ ?
My attempt :
- By using the mean value theorem, for all $(u, v) \in \mathbb{R}^{2},$ we have $$ |\sin (u)-\sin (v)| \leq|u-v| \text { et }|\arctan (u)-\arctan (v)| \leq|u-v| $$ so, $$ \begin{aligned} \left\|f(x, y)-f\left(x^{\prime}, y^{\prime}\right)\right\|_{1} & \leq \frac{1}{4}\left|\sin (x+y)-\sin \left(x^{\prime}+y^{\prime}\right)\right|+\frac{2}{3}\left|\arctan (x-y)-\arctan \left(x^{\prime}-y^{\prime}\right)\right| \\ & \leq \frac{1}{4}\left|(x+y)-\left(x^{\prime}+y^{\prime}\right)\right|+\frac{2}{3}\left|(x-y)-\left(x^{\prime}-y^{\prime}\right)\right| \\ & \leq \frac{1}{4}\left|\left(x-x^{\prime}\right)+\left(y-y^{\prime}\right)\right|+\frac{2}{3}\left|\left(x-x^{\prime}\right)-\left(y-y^{\prime}\right)\right| \\ & \leq\left(\frac{1}{4}+\frac{2}{3}\right)\left(\left|x-x^{\prime}\right|+\left|y-y^{\prime}\right|\right) \\ & \leq \frac{11}{12}\left\|(x, y)-\left(x^{\prime}, y^{\prime}\right)\right\|_{1} \end{aligned} $$
- The function $f$ is contactant, and $(\mathbb{R}^{2},\|.\|_{1})$ is a Banach space. According to the fixed point theorem, there exists a unique couple $(x, y) \in \mathbb{R}^{2}$ such that $(x, y)=f(x, y) .$ In other words, the system admits a unique solution.
- I got stuck here, so any help is appreciated !